Menu Close

5-x-1-2-625-5-x-2-2-5-2x-3-




Question Number 84986 by jagoll last updated on 18/Mar/20
5^((x+1)^2 )  + 625 ≤ 5^(x^2 +2)  + 5^(2x+3)
$$\mathrm{5}^{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \:+\:\mathrm{625}\:\leqslant\:\mathrm{5}^{\mathrm{x}^{\mathrm{2}} +\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2x}+\mathrm{3}} \: \\ $$
Commented by mr W last updated on 18/Mar/20
x≤−(√2) ∨ (1/2)≤x≤(√2)
$${x}\leqslant−\sqrt{\mathrm{2}}\:\vee\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\sqrt{\mathrm{2}} \\ $$
Answered by john santu last updated on 18/Mar/20
5^(x^2 +2x+1)  +625 ≤ 25.5^x^2   + 25.5^(2x+1)   5^x^2   .5^(2x+1)  +625 ≤ 25.5^x^2   + 25.5^(2x+1)   let u = 5^x^2   & v = 5^(2x+1)   uv + 625 ≤ 25u + 25v  uv −25u −25v + 625 ≤ 0  u(v−25)−25(v−25) ≤ 0  (u−25)(v−25)≤ 0  case 1. 5^x^2   ≥ 5^2  ∧ 5^(2x+1)  ≤ 5^2   x≤ −(√2) ∨x≥(√2) ∧ x≤ (1/2)  ∴ x ≤ −(√2)  case 2. ⇒5^x^2   ≤ 5^2  ∧ 5^(2x+1)  ≥5^2   −(√2) ≤ x ≤ (√2) ∧ x ≥ (1/2)  ∴ (1/2)≤x≤(√2)  hence the solution is   x ≤ −(√2) ∨ (1/2)≤x≤(√2)
$$\mathrm{5}^{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}} \:+\mathrm{625}\:\leqslant\:\mathrm{25}.\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:+\:\mathrm{25}.\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \\ $$$$\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:.\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \:+\mathrm{625}\:\leqslant\:\mathrm{25}.\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:+\:\mathrm{25}.\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:\&\:\mathrm{v}\:=\:\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \\ $$$$\mathrm{uv}\:+\:\mathrm{625}\:\leqslant\:\mathrm{25u}\:+\:\mathrm{25v} \\ $$$$\mathrm{uv}\:−\mathrm{25u}\:−\mathrm{25v}\:+\:\mathrm{625}\:\leqslant\:\mathrm{0} \\ $$$$\mathrm{u}\left(\mathrm{v}−\mathrm{25}\right)−\mathrm{25}\left(\mathrm{v}−\mathrm{25}\right)\:\leqslant\:\mathrm{0} \\ $$$$\left(\mathrm{u}−\mathrm{25}\right)\left(\mathrm{v}−\mathrm{25}\right)\leqslant\:\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{1}.\:\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:\geqslant\:\mathrm{5}^{\mathrm{2}} \:\wedge\:\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \:\leqslant\:\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{x}\leqslant\:−\sqrt{\mathrm{2}}\:\vee\mathrm{x}\geqslant\sqrt{\mathrm{2}}\:\wedge\:\mathrm{x}\leqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:\mathrm{x}\:\leqslant\:−\sqrt{\mathrm{2}} \\ $$$$\mathrm{case}\:\mathrm{2}.\:\Rightarrow\mathrm{5}^{\mathrm{x}^{\mathrm{2}} } \:\leqslant\:\mathrm{5}^{\mathrm{2}} \:\wedge\:\mathrm{5}^{\mathrm{2x}+\mathrm{1}} \:\geqslant\mathrm{5}^{\mathrm{2}} \\ $$$$−\sqrt{\mathrm{2}}\:\leqslant\:\mathrm{x}\:\leqslant\:\sqrt{\mathrm{2}}\:\wedge\:\mathrm{x}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\sqrt{\mathrm{2}} \\ $$$$\mathrm{hence}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\: \\ $$$$\mathrm{x}\:\leqslant\:−\sqrt{\mathrm{2}}\:\vee\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\sqrt{\mathrm{2}} \\ $$
Commented by jagoll last updated on 18/Mar/20
waw..mr w & mr john thanks
$$\mathrm{waw}..\mathrm{mr}\:\mathrm{w}\:\&\:\mathrm{mr}\:\mathrm{john}\:\mathrm{thanks} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *